Question

A hanging mass, \(M_{1}=0.500 \mathrm{~kg}\), is attached by a light string that runs over a frictionless pulley to the front of a mass \(M_{2}=1.50 \mathrm{~kg}\) that is initially at rest on a frictionless table. A third mass \(M_{3}=2.50 \mathrm{~kg}\), which is also initially at rest on the same frictionless table, is attached to the back of \(M_{2}\) by a light string. a) Find the magnitude of the acceleration, \(a,\) of mass \(M_{3}\). b) Find the tension in the string between masses \(M_{1}\) and \(M_{2}\).

Short answer

In this problem, we have three masses connected by two strings. Mass M1 is hanging off the edge of the table, M2 is on the table, and M3 is connected to M2. To find the acceleration of mass M3 and the tension in the string between masses M1 and M2, we first identified the forces acting on each mass and wrote the equations of motion using Newton's second law. We then eliminated the tensions and solved for the acceleration, which we found to be approximately -1.31 m/s^2 (magnitude: 1.31 m/s^2). Finally, we calculated the tension in the string between masses M1 and M2, which was approximately 4.25 N.

Step-by-step solution

Identify forces acting on each mass

For mass M1, there are two forces acting: the gravitational force and the tension in the string. For mass M2, there are two forces acting: the tension in the string between M1 and M2 and the tension in the string between M2 and M3. For mass M3, there is only one force acting: the tension in the string between M2 and M3.

Write equations of motion using Newton's second law

We can write the equations of motion using the form \(F_{net} = ma\). For mass M1: \(T_1 - M_1g = -M_1a\) For mass M2: \(T_1 - T_2 = M_2a\) For mass M3: \(T_2 = M_3a\)

Find the acceleration of mass M3

We can eliminate the tensions T1 and T2 from the equations and solve for the acceleration, a:\ Add the equations for masses M1 and M2: \( T_1 - M_1g + T_1 - T_2 = -M_1a + M_2a \Rightarrow 2T_1 - M_1g - T_2 = (M_2 - M_1)a\) Now, substitute \(T_2 = M_3a\) from the equation for mass M3: \( 2T_1 - M_1g - M_3a = (M_2 - M_1)a\) Substitute \(T_1 = M_1a + M_1g\) from the equation for mass M1: \( 2(M_1a + M_1g) - M_1g - M_3a = (M_2 - M_1)a\) Simplify the equation and solve for a: \(a(2M_1 - M_3 - M_1 + M_1) = M_1g\) \(a=\frac{M_1g}{2M_1 - M_3} = \frac{(0.5)(9.81)}{2(0.5) - 2.5} \Rightarrow a \approx -1.31 \frac{m}{s^2}\) (with a negative sign because it is in the opposite direction of M1 tension force) The magnitude of the acceleration, \(a\), of mass \(M_3\) is 1.31 \(\frac{m}{s^2}\).

Find the tension in the string between masses M1 and M2

We can use the equation of motion for mass M1 to find the tension T1: \(T_1 = M_1a + M_1g = (0.5)(-1.31) + (0.5)(9.81) \Rightarrow\) \(T_1 \approx 4.25 N\) The tension in the string between masses \(M_{1}\) and \(M_{2}\) is 4.25 N.

Key concepts

Newton's Second Law

Newton's second law of motion is crucial for understanding the dynamics of objects. It states that the net force acting on an object is equal to the mass of the object times its acceleration \( F = ma \). This principle is a cornerstone in physics problem solving, particularly in exercises involving multiple mass systems and forces.

Applying Newton's second law to our textbook example, we start by identifying the forces and then construct equations for each mass. For instance, the mass hanging from the string experiences both its weight \( M_1g \) — the gravitational force pulling it downwards — and the tension in the string pulling upwards. Newton's second law allows us to set up a relationship between these forces, the mass \( M_1 \), and its acceleration \( a \).

Equations of Motion

In addressing physics problems, equations of motion are used to describe how an object moves in terms of its velocity, acceleration, and the forces acting upon it. Our example makes use of these equations to relate the tension in the strings and the gravitational force to the acceleration of the masses. After establishing the net forces for each mass, we utilize algebraic manipulation to solve for the desired quantities:

- For mass \( M_1 \), \( T_1 - M_1g = -M_1a \)
- For mass \( M_2 \), \( T_1 - T_2 = M_2a \)
- For mass \( M_3 \), \( T_2 = M_3a \)

These equations encapsulate the aforementioned Newton's second law and allow us to compute the acceleration and tensions by substituting known values and solving for the unknowns.

Tension in a String

Understanding the concept of tension is critical when dealing with physics problems involving strings and cables. Tension is the pulling force transmitted along the length of a flexible connector, such as a string or cable. In our exercise, tension plays a key role in transferring the force from one mass to another across a pulley system.

To compute the tension, we need to analyze the forces acting on the individual masses and apply Newton's laws. For instance, tension \(T_1\) is what pulls mass \(M_2\) across the table, and it is also what balances the gravitational force acting on mass \(M_1\) to set it into motion. Similarly, tension \(T_2\) acts on mass \(M_3\) alone, and its value can be deduced from the equation \(T_2 = M_3a\). Clearly, the tension in the strings is an essential factor in how the system behaves and is directly related to the acceleration of the masses involved.

Gravitational Force

Gravitational force, also known as weight, is the force with which a planet, such as the Earth, pulls objects towards its center. In physics problems, it is denoted by \( F_g \), or more commonly by \( mg \), where \( m \) is the object's mass and \( g \) is the acceleration due to gravity (approximately \( 9.81 m/s^2 \) on Earth).

In the given problem, gravitational force acts on mass \(M_1\) and is the driving force that initiates the motion of the connected mass system. This force is essential in determining the acceleration of the masses, as seen in the solution where the value of \( g \) is used to solve for the acceleration \( a \) and to find tension \( T_1 \) between masses \(M_1\) and \(M_2\). Thus, gravitational force not only acts on mass \(M_1\) but influences the entire chain of connected masses in this physics scenario.